Lieb-Liniger equations

A gas of repulsive bosons interacting via $\delta$-potential in one-dimensional system is described by the Hamiltonian (5.1) with the relation between the coupling constant $g_{1D}$ and the scattering length $a_{1D}$ given by (1.69). It was shown by Lieb and Liniger [LL63] that the ground state energy can be found from the solution of the integral equation (see for example, [LL63,DLO01]):

$$\rho(k) = \frac{1}{2\pi} +\int\limits_{-1}^1\frac{2\lambda\rho(\varkappa)}{\lambda^2+(k-\varkappa)^2}\,\frac{d\varkappa}{2\pi}$$ (10.1)

Normalization of the function $\rho(k)$ is related to the density $n\vert a_{1D}\vert$:

$$\gamma = \frac{2}{n\vert a_{1D}\vert} = \int\limits_{-1}^1 \rho(k)\,dk,$$ (10.2)

here we use parameter $\gamma$ which is often introduced for solving the Bethe equations and is inversely proportional to the density.

The energy per particle $E/N = n^2 e(n\vert a_{1D}\vert) \hbar^2/2m$ is obtained from integral

$$e(n\vert a_{1D}\vert) = %%\left(\frac{2}{\lambda\, n\vert a_{1D}\vert}\right)^3 \frac{\gamma^3}{\lambda^3}\int\limits_{-1}^1 k^2\rho(k)\,dk$$ (10.3)

The procedure of solving the integral equations can be following:

1. Fix some value of $\lambda$
2. Obtain $\rho(k)$ from (A.1)
3. Obtain density $n\vert a_{1D}\vert$ from (A.2)
4. Obtain energy $e(n\vert a_{1D}\vert)$ from (A.3)